Is there a limitation of Assuming that I am not aware of? I would like to assume that the square of a number is negative:
Assuming[x^2 < 0, Simplify[Sqrt[x^2]]]
x
Why is $Assumptions generating a message that x^2 < 0 is a contradictory assumption?
The Assumptions documentation states that
Quantities that appear algebraically in inequalities are always assumed to be real.
Thus in x^2 < 0 the variable x is assumed to be real. That is why the assumption is contradictory.
A shorter example is
$Assumptions = x^2 < 0
During evaluation of $Assumptions::cas: Warning: contradictory assumption(s) x^2<0 encountered.
(* x^2 < 0 *)
Other functions, such as Reduce, also make similar automatic assumptions. This is also documented for Reduce.
Reduce[x^2 < 0, x]
(* False *)
However, with Reduce, it is possible to override this automatic assumption.
Reduce[x^2 < 0, x, Complexes]
(* Re[x] == 0 && (Im[x] < 0 || Im[x] > 0) *)
I do not know if this is possible with Assumptions.
Reduce for $Assumptions? Or one could use x^2 \[Element] Reals && Re[x^2] < 0 (the Re seems to keep $Assumptions from complaining).
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Commented
Nov 23, 2018 at 20:22
Sqrt[x^2] can't be simplified to x correctly with these assumptions, e.g. (-I)^2 == I^2 == -1.
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Im[x] > 0 (resp. Im[x] < 0), then Simplify will return x (resp. -x). It's not clear to me that getting x is the desired outcome, though.
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Commented
Nov 23, 2018 at 21:31
$Assumptions = x^2 < 0$\endgroup$Reduce[x^2 < 0, x]-->Falseand the statement in the documentation that "Reduce assumes by default that quantities appearing algebraically in inequalities are real, while all other quantities are complex." The assumption can be overridden.Reduce[x^2 < 0, x, Complexes]$\endgroup$